A 4.850 g sample containing KCl and KClO 4 was dissolved in sufficient water to

Question

A 4.850 g sample containing KCl and KClO4 was dissolved in sufficient water to give 250.00 mL of solution. A 50.00 mL portion of the solution required 42.75 mL of 0.0750 M AgNO3 in a Mohr titration. Next, a 25.00 mL portion of the original solution was treated with V2(SO4)3 to reduce the perchlorate ion to chloride,

8V3+(aq) + ClO4-(aq) + 12H2O(?)

? Cl-(aq) + 8VO2+(aq) + 8H3O+(aq)

and the resulting solution was titrated with AgNO3. This titration required 38.39 mL of 0.0750 M AgNO3. What is the mass percent of KCl and KClO4 in the mixture?

% KCl

% KClO4

A 4.850 g sample containing KCl and KClO4 was dissolved in sufficient water to give 250.00 mL of solution. A 50.00 mL portion of the solution required 42.75 mL of 0.0750 M AgNO3 in a Mohr titration. Next, a 25.00 mL portion of the original solution was treated with V2(SO4)3 to reduce the perchlorate ion to chloride, 8V3+(aq) + ClO4-(aq) + 12H2O(??) ?? Cl-(aq) + 8VO2+(aq) + 8H3O+(aq) and the resulting solution was titrated with AgNO3. This titration required 38.39 mL of 0.0750 M AgNO3. What is the mass percent of KCl and KClO4 in the mixture? % KCl % KClO4

Explanation / Answer

Sample weight = 4.850 g

solution made in 250 ml = 0.250 L

1) 50 ml aliquot analysis

moles of AgNO3 used = 0.0750 x 0.04275 = 3.206 x 10^-3 mols

moles of Cl- = 3.206 x 10^-3 mols

Molarity of [Cl-] = 3.206 x 10^-3/(0.050 + 0.04275) = 0.0346 M

moles of [Cl-] in original solution = 0.0346 x 0.250 = 8.65 x 10^-3 mols

mass of KCl = 8.65 x 10^-3 x 74.5513 = 0.645 g

% of KCl in sample = (0.645/4.850) x 100 = 13.30%

2) 25 ml aliquot analysis

AgNO3 used = 0.0750 x 0.03839 = 2.88 x 10^-3 mols

moles of ClO4- = 2.88 x 10^-3 mols

concentration of [ClO4-] = 2.88 x 10^-3/(0.03839 + 0.025) = 0.045 M

moles of [ClO4-] in sample = 0.045 x 0.250 = 0.01136 mols

mass of KClO4 = 0.01136 x 138.55 = 1.574 g

mass % for KClO4 = (1.574/4.850) x 100 = 32.45%

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